Question

Решите уравнение !!!!!!!!!

㏒₃x - ㏒₃(x² - 6)=0

Answer 1

ОДЗ: 1) x > 0               2) x² - 6 > 0
                                        (x - √6 )(x + √6 ) > 0
                                             +                -                +
                                        _______________________
                                                     -√6               √6
ОДЗ : x ∈ ( √6 ; + ∞)      

$log _{3}x-log _{3} ( x^{2} -6)=0\\\\log _{3} \frac{x}{ x^{2} -6} =0\\\\ \frac{x}{ x^{2} -6} =1\\\\ \frac{x}{ x^{2} -6} -1=0\\\\ \frac{x- x^{2} +6}{ x^{2} -6} =0$

- x² + x + 6 = 0                 x² - 6 ≠ 0
x² - x - 6 = 0
D = (-1)² - 4 * 1 * (- 6) = 1+24 = 25 = 5²
$x _{1} = \frac{1+5}{2} =3\\x _{2}= \frac{1-5}{2}= - 2$
x = - 2 - не подходит
Ответ: 3

Answer 2

$log_3x-log_3(x^2-6)=0\; ,\\\\ODZ:\; \left \{ {{x\ \textgreater \ 0} \atop {x^2-6\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 0} \atop {(x-\sqrt6)(x+\sqrt6)\ \textgreater \ 0}} \right. \; \; \to \; \; x\in (\sqrt6,+\infty )\\\\log_3x=log_3(x^2-6)\; \; \Rightarrow \; \; \; x=x^2-6\\\\x^2-x-6=0\\\\x_1=-2\notin ODZ\; ,\; \; x_2=3\; \; \; (teorema\; Vieta)\\\\Otvet:\; \; x=3\; .$